Question about proof from a guy with a highschool education

[reenmachine]

$$G\circ F=\left\{(x,z)\,|\,\exists y\in B~~ \left((x,y)\in F~\land~(y,z)\in G\right)\right\}.$$

What do you think of this alternative statement?

∀x,y,z ( (¬(F(x)∧ ¬y)∧¬(G(y)∧ ¬z)) ∧ (¬(G(F(x)) ∧ ¬z) )

Edit second try:

∀x,y,z ( (¬(x ∈ F ∧ ¬y ∈ F)∧ ¬(y ∈ G ∧ ¬z ∈ G)) ∧ (¬(x ∈ F ∧ ¬z ∈ G) )

 

[Fredrik]

edit: Just for the sake of ultimate clearness , and I know we already talked about this , but in the last set notation , ''for each'' (x,z) is implied on the left side correct? Don't need to write the symbol?

Right. You just need to keep in mind that for all properties P, the notation $\{x\,|\,P(x)\}$ is read as "the set of all all x such that P(x)" or (equivalently) "the set of all x with property P".

What do you think of this alternative statement?

∀x,y,z ( (¬(F(x)∧ ¬y)∧¬(G(y)∧ ¬z)) ∧ (¬(G(F(x)) ∧ ¬z) )

Edit second try:

∀x,y,z ( (¬(x ∈ F ∧ ¬y ∈ F)∧ ¬(y ∈ G ∧ ¬z ∈ G)) ∧ (¬(x ∈ F ∧ ¬z ∈ G) )

In the first one, "$\lnot y$" doesn't make sense, since y is a set, not a statement about sets.

The second one is equivalent to
$$\forall x,y,z~\left((x\in F\Rightarrow y\notin F)\land\dots\right).$$ It's clearly not true for all sets x and y that if x is in F, then y is not. Edit: This is wrong. It should be $y\in F$ where I said $y\notin F$. So the sentence above should be "It's clearly not true for all sets x and y that if x is in F, then so is y.".

If you meant $$\{(x,z)\,|\, \text{your statement}\},$$ then you missed the fact that variables that are targets of a "for all" are dummy variables, so your x,y,z don't have anything to do with the x and z before the | symbol.

 

[reenmachine]

Right. You just need to keep in mind that for all properties P, the notation $\{x\,|\,P(x)\}$ is read as "the set of all all x such that P(x)" or (equivalently) "the set of all x with property P".

This is what I thought thank you!

In the first one, "$\lnot y$" doesn't make sense, since y is a set, not a statement about sets.

Make sense.

The second one is equivalent to
$$\forall x,y,z~\left((x\in F\Rightarrow y\notin F)\land\dots\right).$$ It's clearly not true for all sets x and y that if x is in F, then y is not.

In the second one I wrote ¬(x ∈ F ∧ ¬y ∈ F) which should have been ¬(x ∈ F ∧ ¬(y ∈ F)) , so I'm not sure why you're saying that x in F implies that y isn't in F.What I meant by this is that:

It's not true that: x is in F AND that it's not true that y is in F.Therefore if x is in F then so is y (at least that's what make sense in my mind based on how I operated using these symbols earlier in the thread).

If you meant $$\{(x,z)\,|\, \text{your statement}\},$$ then you missed the fact that variables that are targets of a "for all" are dummy variables, so your x,y,z don't have anything to do with the x and z before the | symbol.

oops

So my statements using the true/false method wouldn't actually require dummy variables? I understand now that there's a difference between dummy variables and the precise x which is in A and so on...but I fail to see how to use these dummy variable to make a notation using the true/false logic method.

thanks again

 

[Fredrik]

In the second one I wrote ¬(x ∈ F ∧ ¬y ∈ F) which should have been ¬(x ∈ F ∧ ¬(y ∈ F)) , so I'm not sure why you're saying that x in F implies that y isn't in F.

$p\rightarrow q$ has the same truth table as $\lnot (p\land\lnot q)$. (This is a good exercise). So these statements are equivalent.

So my statements using the true/false method wouldn't actually require dummy variables? I understand now that there's a difference between dummy variables and the precise x which is in A but f(x)=y and so on...but I fail to see how to use these dummy variable to make a notation using the true/false logic method.

It's hard to explain, since I don't see what you're trying to do.

Note by the way that none of the x,y,z in my notation can be a member of F. They are elements of A,B,C respectively, and F is a subset of A×B.

 

[reenmachine]

$p\rightarrow q$ has the same truth table as $\lnot (p\land\lnot q)$. (This is a good exercise). So these statements are equivalent.It's hard to explain, since I don't see what you're trying to do.

OKay , but in that case if

p= x ∈ F and q= y ∈ F

then x ∈ F → y ∈ F = ¬(x ∈ F ∧ ¬y ∈ F)?

I'm a little bit confused where my statement indicate that if x is in F then y isn't in F based on your following statement:

$$\forall x,y,z~\left((x\in F\Rightarrow y\notin F)\land\dots\right).$$

Note by the way that none of the x,y,z in my notation can be a member of F. They are elements of A,B,C respectively, and F is a subset of A×B.

So G is a subset of B×C and G°F is a subset of A×C?

 

[Fredrik]

OKay , but in that case if

p= x ∈ F and q= y ∈ F

then x ∈ F → y ∈ F = ¬(x ∈ F ∧ ¬y ∈ F)?

Yes, $p\rightarrow q$ is equivalent to $\lnot (p\land\lnot q)$ for all p and q (as proved by their truth tables). This specific choice of p and q is not an exception.

(I would however write $\leftrightarrow$ or $\Leftrightarrow$ instead of =).

I'm a little bit confused where my statement indicate that if x is in F then y isn't in F

You said $\lnot (x\in F\land\lnot y\in F)$, which is equivalent to $x\in F\rightarrow y\in F$. So I should have said "if x is in F, then y is in F".

So G is a subset of B×C and G°F is a subset of A×C?

Yes.

 

[reenmachine]

Let me try it one more time.

∀x,y,z ( ¬(x ∈ A ∧ ¬y ∈ B) ∧ ¬(y ∈ B ∧ ¬z ∈ C) )

I am trying to describe or define G°F.

Yes, $p\rightarrow q$ is equivalent to $\lnot (p\land\lnot q)$ for all p and q (as proved by their truth tables). This specific choice of p and q is not an exception.

(I would however write $\leftrightarrow$ or $\Leftrightarrow$ instead of =).

very clear

You said $\lnot (x\in F\land\lnot y\in F)$, which is equivalent to $x\in F\rightarrow y\in F$. So I should have said "if x is in F, then y is in F".

good

thanks a lot !

 

[CompuChip]

I would define $G\circ F:A\to C$ by saying that
$$(G\circ F)(x)=G(F(x))$$ for all $x\in A$. I have never thought about other ways to define it until now, but let's see.

Can't you just use the definition of a function we have been using all the time to define
(x, z) \in G \circ F \subseteq A \times C \iff x \in A \land \exists y \in B \text{ s.t. } ( (x, y) \in F \land (y, z) \in G)
where F \subseteq A \times B and G \subseteq B \times C
(note that I'm using numerous notational "shortcuts" here to make it readable).

 

[Fredrik]

Let me try it one more time.

∀x,y,z ( ¬(x ∈ A ∧ ¬y ∈ B) ∧ ¬(y ∈ B ∧ ¬z ∈ C) )

I am trying to describe or define G°F.

And this is only part of the notation? The whole thing is $\{(x,z)\,|\,\text{your statement}\}$? In that case, it still has the problem that the x,y,z have nothing to do with the (x,z). If your statement is true, the set will be "the set of all ordered pairs" (taken from any two sets), which I'm sure is too large to even exist in ZFC set theory. If your statement is false, the set will be ∅.

The part

¬(x ∈ A ∧ ¬y ∈ B) ∧ ¬(y ∈ B ∧ ¬z ∈ C)​

says that if x is in A then y is in B, and if y is in B then z is in C. This is not true for all x,y,z.

 

[Fredrik]

Can't you just use the definition of a function we have been using all the time to define
(x, z) \in G \circ F \subseteq A \times C \iff x \in A \land \exists y \in B \text{ s.t. } ( (x, y) \in F \land (y, z) \in G)
where F \subseteq A \times B and G \subseteq B \times C
(note that I'm using numerous notational "shortcuts" here to make it readable).

Yes, but I was trying to explain why the more familiar definition of $G\circ F$ implies that $G\circ F$ is the set specified in the text.

 

[CompuChip]

Ah, right. Sometimes there are so many new posts in a few hours (or a day) that I fail to read them all :)

 

[reenmachine]

And this is only part of the notation? The whole thing is $\{(x,z)\,|\,\text{your statement}\}$? In that case, it still has the problem that the x,y,z have nothing to do with the (x,z). If your statement is true, the set will be "the set of all ordered pairs" (taken from any two sets), which I'm sure is too large to even exist in ZFC set theory. If your statement is false, the set will be ∅.

The part

¬(x ∈ A ∧ ¬y ∈ B) ∧ ¬(y ∈ B ∧ ¬z ∈ C)​

says that if x is in A then y is in B, and if y is in B then z is in C. This is not true for all x,y,z.

Ok , I took the liberty to try my luck at defining G°F using the less shortcuts possible but it seems I'm having problems.It's not the end of the world since I do understand how to define it ''normally''.

thanks man!

 

[reenmachine]

In the book of proof they introduced the concept of indexed sets.

From what I understood of it , $A_{1} \cup A_{2} \cup A_{3} \cup ... \cup A_{n}$ means that $x \in A_{i}$ in at least one of the set $A_{i}$.

If you replace the $\cup$ with the $\cap$ , then $x \in A_{i}$ for every set $A_{i}$.

$$\bigcup_{i=1}^n A_i = A_{1} \cup A_{2} \cup ... \cup A_{n}$$

The same is true if you replace the $\cup$ with a $\cap$.If there's a known number of set , you could just replace n with that number on top of the symbol.

They talked about the set I , which is called an index set.They say that $i \in I$ .What exactly is the purpose of this set?

Suppose we have $$\bigcup_{i=1}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$ does it mean that $1 \in I$ , $2 \in I$ and $3 \in I$?thanks ! trying my best to get used to LaTeX , not sure how the post will come out.

 

[Fredrik]

In the book of proof they introduced the concept of indexed sets.

From what I understood of it , $A_{1} \cup A_{2} \cup A_{3} \cup ... \cup A_{n}$ means that $x \in A_{i}$ in at least one of the set $A_{i}$.

You left out the $x\in$ at the beginning. You can type \dots or \cdots instead of ...

If you replace the $\cup$ with the $\cap$ , then $x \in A_{i}$ for every set $A_{i}$.

Yes.

They talked about the set I , which is called an index set.They say that $i \in I$ .What exactly is the purpose of this set?

It gives us more options with the notation. For example, you can write $\{A_i|i\in I\}$ instead of $\{A_1,\dots,A_n\}$. Instead of ending a sentence with

...for all $x\in\{A_1,\dots,A_n\}$.​

you can end it with

...for all $A_i$ with $i\in I$.​

Suppose we have $$\bigcup_{i=1}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$ does it mean that $1 \in I$ , $2 \in I$ and $3 \in I$?

The symbol $I$ isn't assigned a meaning by the above. You have to explicitly say that $I=\{1,2,3\}$ before you can say that $1,2,3\in I$.

trying my best to get used to LaTeX , not sure how the post will come out.

Looks pretty good. No need to post your experiments, since we have a preview feature. I used my new superpowers to delete your previous post, as requested.

 

[reenmachine]

You left out the $x\in$ at the beginning.

Do you mean this:

$x\in A_{1} \cup A_{2} \cup A_{3} \cup ... \cup A_{n}$ means that $x \in A_{i}$ in at least one of the set $A_{i}$ ?

It gives us more options with the notation. For example, you can write $\{A_i|i\in I\}$ instead of $\{A_1,\dots,A_n\}$. Instead of ending a sentence with

...for all $x\in\{A_1,\dots,A_n\}$.​

you can end it with

...for all $A_i$ with $i\in I$.​

Hmm ok , but in this case set $I$ would be the set of all numbers in $A_{1} , A_{2} , ... , A_{n}$ so $I = \{1 , 2 , \dots , n\}$ ?

The symbol $I$ isn't assigned a meaning by the above. You have to explicitly say that $I=\{1,2,3\}$ before you can say that $1,2,3\in I$.

Would this do the trick?

$$\bigcup_{i \in I}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$
or
$$\bigcup_{i = I}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$

So $I=\{1,2,3\}$ (in that case)

Looks pretty good. No need to post your experiments, since we have a preview feature. I used my new superpowers to delete your previous post, as requested.

Thanks , I'll try to use LaTeX more and more from now on.I will often have to edit my posts though.

 

[Fredrik]

Do you mean this:

$x\in A_{1} \cup A_{2} \cup A_{3} \cup ... \cup A_{n}$ means that $x \in A_{i}$ in at least one of the set $A_{i}$ ?

Yes. You could also end the sentence with "...means that $x\in A_i$ for at least one $i\in\{1,\dots,n\}$".

Note that for all x,
\begin{align}
x\in\bigcup_{i\in I}A_i\ \Leftrightarrow\ \exists i\in I~~x\in A_i,\\
x\in\bigcap_{i\in I}A_i\ \Leftrightarrow\ \forall i\in I~~x\in A_i.
\end{align} What I mean by $\exists i\in I~~x\in A_i$ is "there's an i in I such that x is in A_i". This can also be written as $\exists i\ \left(i\in I\ \land\ x\in A_i\right)$. Edit: Nope, that last statement is nonsense.

Hmm ok , but in this case set $I$ would be the set of all numbers in $A_{1} , A_{2} , ... , A_{n}$ so $I = \{1 , 2 , \dots , n\}$ ?

Yes again.

Would this do the trick?
$$\bigcup_{i \in I}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$

Not really, because what if $A_4=A_1\cup A_2\cup A_3$? Then we also have $\bigcup_{i\in I} A_i=A_4$, suggesting that $I=\{4\}$. Edit: I didn't notice that you put a 3 on top of the $\bigcup$. That's not a standard notation.

 

[reenmachine]

Yes. You could also end the sentence with "...means that $x\in A_i$ for at least one $i\in\{1,\dots,n\}$".

Note that for all x,
\begin{align}
x\in\bigcup_{i\in I}A_i\ \Leftrightarrow\ \exists i\in I~~x\in A_i,\\
x\in\bigcap_{i\in I}A_i\ \Leftrightarrow\ \forall i\in I~~x\in A_i.
\end{align} What I mean by $\exists i\in I~~x\in A_i$ is "there's an i in I such that x is in A_i". This can also be written as $\exists i\ \left(i\in I\ \land\ x\in A_i\right)$.

Okay so I could also write $\forall i\ \left(i\in I\ \land\ x\in A_i\right)$ to define the second one?

Not really, because what if $A_4=A_1\cup A_2\cup A_3$? Then we also have $\bigcup_{i\in I} A_i=A_4$, suggesting that $I=\{4\}$.

What about this then?
$$\bigcup_{i = \{1,2,3\}}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$

If not , I'm not sure what to do.Is it useless to try inserting the definition of $I$ in the notation or should I simply define it separately before starting to define some other set that $I=\{1,2,3\}$ ?

EDIT: What I mean by this is doing something like:
Set $I = \{1,2,3\}$
$$\bigcup_{i \in I}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$

...to describe the set $\{A_{1} \cup A_{2} \cup A_{3}\}$?

It's not clear what $I$ is unless I previously defined it.

thanks!

 

[Fredrik]

Looks like I did that too quickly. The statement $\forall i\ \left(i\in I\ \land\ x\in A_i\right)$ would imply that I is a set that contains all sets, so it's nonsense.

 

[reenmachine]

Looks like I did that too quickly. The statement $\forall i\ \left(i\in I\ \land\ x\in A_i\right)$ would imply that I is a set that contains all sets, so it's nonsense.

What do you mean? Did you made a mistake in your previous post? (just to be sure we're on the same page)

 

[micromass]

This can also be written as $\exists i\ \left(i\in I\ \land\ x\in A_i\right)$. Edit: Nope, that last statement is nonsense.

I'm not sure why you think that this statement is nonsense. Something like $\exists x:~P(x)$ or $\forall x:~P(x)$ is perfectly allowed. Quantifiers can take as range all sets. The only thing that's not allowed is to call $\{x~\vert~P(x)\}$ a set.

 

[Fredrik]

I'm busy now, I'll explain what I was thinking later.

 

[reenmachine]

The only thing that's not allowed is to call $\{x~\vert~P(x)\}$ a set.

While we're talking about this might as well refresh my mind.

Suppose the set $A = \{x~\vert~x \in A\}$ , this would be an incorrect definition because it says nothing about A except that A is made of his own elements?

Something like $A = \{x \in A~\vert~x\not \in B\}$ if $A = B'$ would be better because it explains what A is is that correct? But I can't say that $A = \{x~\vert~x\not \in B\}$.

Why can't I say $A = \{x~\vert~x\in A \land x\not\in B\}$ ? Is it because I can't use A in it's own definition , a little bit like not using a word in it's own definition?

thanks

 

[reenmachine]

I'm busy now, I'll explain what I was thinking later.

No problem man! Take all the time you need , I'm already considering myself very lucky to receive so many good answers to my questions!

 

[Fredrik]

What do you mean? Did you made a mistake in your previous post? (just to be sure we're on the same page)

Yes, that's what I meant. This statement is OK:

Note that for all x,
\begin{align}
x\in\bigcup_{i\in I}A_i\ \Leftrightarrow\ \exists i\in I~~x\in A_i,\\
x\in\bigcap_{i\in I}A_i\ \Leftrightarrow\ \forall i\in I~~x\in A_i.
\end{align}

This one is not:

This can also be written as $\exists i\ \left(i\in I\ \land\ x\in A_i\right)$.

I'm not sure why you think that this statement is nonsense. Something like $\exists x:~P(x)$ or $\forall x:~P(x)$ is perfectly allowed. Quantifiers can take as range all sets. The only thing that's not allowed is to call $\{x~\vert~P(x)\}$ a set.

Yes, I understand that, but I said
$$\forall i~~ (i\in I\ \land\ \text{something}),$$ and didn't say anything to restrict the scope of the $\forall$. So the sentence says that all sets in ZFC set theory are members of the set I.

@reenmachine: The statement "for all i in I, x is in A_i" can be written as
$$\forall i\in I~~ x\in A_i.$$ This should be viewed as a simplified notation for the following sentence:
$$\forall i~~\left(i\in I\ \Rightarrow x\in A_i\right).$$ My mistake was to write $\land$ instead of $\Rightarrow$.

 

[Fredrik]

Okay so I could also write $\forall i\ \left(i\in I\ \land\ x\in A_i\right)$ to define the second one?

Even if we replace $\land$ with $\Rightarrow$, all we have there is a statement about the set x, so it can't by itself define the set $\bigcup_{i\in I} A_i$. It says that for all $i\in I$, we have $x\in A_i$. This means that x is an element of every $A_i$ with $i\in I$. A statement like that must be a part of the definition, but it can't be the whole thing.

What about this then?
$$\bigcup_{i = \{1,2,3\}}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$

There are only two standard notations,
$$\bigcup_{i\in I}A_i,$$ and $$\bigcup_{i=1}^n A_i.$$

If not , I'm not sure what to do.Is it useless to try inserting the definition of $I$ in the notation or should I simply define it separately before starting to define some other set that $I=\{1,2,3\}$ ?

I'm not sure what you've been trying to do, but yes, if you want to assign a meaning to the symbol I, you have to do it using one of the standard ways to specify a set. In this case, $I=\{1,2,3\}$ is the obvious option, and $I=\{n\in\mathbb Z\,|\,1\leq n\leq 3\}$ is one of the alternatives.

EDIT: What I mean by this is doing something like:
Set $I = \{1,2,3\}$
$$\bigcup_{i \in I}^3 A_i = A_{1} \cup A_{2} \cup A_{3}$$

Lose the 3 on top, and the notation is fine.

...to describe the set $\{A_{1} \cup A_{2} \cup A_{3}\}$?

It doesn't really describe it. It's just an alternative notation for it. (And I think you meant $A_{1} \cup A_{2} \cup A_{3}$).

It's not clear what $I$ is unless I previously defined it.

Exactly.

Suppose the set $A = \{x~\vert~x \in A\}$ , this would be an incorrect definition because it says nothing about A except that A is made of his own elements?

That's right. The statement would be true, but it doesn't tell us anything about the sets represented by the symbols in it (in particular A). Such statements are said to be vacuously true.

Something like $A = \{x \in A~\vert~x\not \in B\}$ if $A = B'$ would be better because it explains what A is is that correct? But I can't say that $A = \{x~\vert~x\not \in B\}$.

B' is only defined when all the sets we're working with are subsets of some set X. Then B' is defined as X-B. B' is not defined here.

But I can't say that $A = \{x~\vert~x\not \in B\}$.

No, but you could say e.g. that $A = \{x\in C~\vert~x\not \in B\}$, if you have already specified what C is.

Why can't I say $A = \{x~\vert~x\in A \land x\not\in B\}$ ? Is it because I can't use A in it's own definition , a little bit like not using a word in it's own definition?

Right, you can't use the definition while you're stating it. But you also need to be careful when you use the notation $\{x\,|\,P(x)\}$, because unlike $\{x\in y\,|\, P(x)\}$, it's not guaranteed by the axioms to always make sense.

 

[reenmachine]

Even if we replace $\land$ with $\Rightarrow$, all we have there is a statement about the set x, so it can't by itself define the set $\bigcup_{i\in I} A_i$. It says that for all $i\in I$, we have $x\in A_i$. This means that x is an element of every $A_i$ with $i\in I$. A statement like that must be a part of the definition, but it can't be the whole thing.

Hmmm , by stating that ''This means that x is an element of every $A_i$ with $i\in I$'' , you are saying that we're making a statement about the set x and not about $\bigcup_{i\in I} A_i$.

I somewhat intuitively understand but still have some confusion.I will take the liberty of creating a scenario to put things into perspective as I think it might help me in that case.Take note that I am thinking as I type the post.

Suppose a room full of people that have either blond , brown or white hair.There's no bald people or people that have another hair color than blond , brown and white.

Set $A_1$ is made of all people in the room with blond hair , set $A_2$ is made of all people in the room with brown hair and set $A_3$ is made of all people in the room with white hair.

Set $I = \{1,2,3\}$

A person with blond hair $\in A_1$ but a person with brown hair $\not \in A_1$ and so on...

We can conclude that the set:
$$\bigcup_{i\in I}A_i = A_1 \cup A_2 \cup A_3$$
...is made of every person in the room.

The earlier statement ''This means that x is an element of every $A_i$ with $i\in I$'' would mean $1,2,3 \in I$ and $x \in A_i$.If he is in $A_i$ and $i \in I$ then $x \in A_1$ , $x \in A_2$ or $x \in A_3$ , but since he either has blond , brown or white hair , how could he be an element of each $A_i$? In a way , to find an element of the big set you have to find what's common between elements from the 3 subsets? Like ''being in the room'' in my previous exemple?

Set notation where x = a person in the room: $\{ x\in\bigcup_{i\in I}A_i : \forall i\in I \land x \in A_i\}$

Does this really define the set of all people in the room?

There are only two standard notations,
$$\bigcup_{i\in I}A_i,$$ and $$\bigcup_{i=1}^n A_i.$$

ok good

I'm not sure what you've been trying to do, but yes, if you want to assign a meaning to the symbol I, you have to do it using one of the standard ways to specify a set. In this case, $I=\{1,2,3\}$ is the obvious option, and $I=\{n\in\mathbb Z\,|\,1\leq n\leq 3\}$ is one of the alternatives.

Ok , so if $I = \{1,2,3,4,5,6,7,8,9\}$ then $I = \{n \in\mathbb Z\,|\,1\leq n\leq 9\}$?

Lose the 3 on top, and the notation is fine.

Ok so we only use the number or symbol on top with we have $i=1$ at the bottom and not $i \in I$ ?

It doesn't really describe it. It's just an alternative notation for it. (And I think you meant $A_{1} \cup A_{2} \cup A_{3}$).

What do you mean? I can't write it between {}?

B' is only defined when all the sets we're working with are subsets of some set X. Then B' is defined as X-B. B' is not defined here.

Ok , so B' doesn't really exist except in naive set theory?

No, but you could say e.g. that $A = \{x\in C~\vert~x\not \in B\}$, if you have already specified what C is.

So I can't say $A = \{x \in A~\vert~x\not \in B\}$ ?

Because of:

Right, you can't use the definition while you're stating it. But you also need to be careful when you use the notation $\{x\,|\,P(x)\}$, because unlike $\{x\in y\,|\, P(x)\}$, it's not guaranteed by the axioms to always make sense.

thanks!

 

[Fredrik]

The earlier statement ''This means that x is an element of every $A_i$ with $i\in I$'' would mean $1,2,3 \in I$ and $x \in A_i$.If he is in $A_i$ and $i \in I$ then $x \in A_1$ , $x \in A_2$ or $x \in A_3$ , but since he either has blond , brown or white hair , how could he be an element of each $A_i$? In a way , to find an element of the big set you have to find what's common between elements from the 3 subsets? Like ''being in the room'' in my previous exemple?

First of all, you shouldn't have included the word "this", because it referred to the previous statement in my post. I assume that you meant that

x is an element of every $A_i$ with $i\in I$ $.\qquad(1)$​

means

$1,2,3 \in I$ and $x \in A_i$ $.\qquad(2)$​

Clearly (2) is very different from (1). (2) tells us that {1,2,3} is a subset of I. (1) tells us nothing about I. (2) tells us that $x\in A_i$. (What is i here?) (1) says that x is in all the A_i.

Set notation where x = a person in the room: $\{ x\in\bigcup_{i\in I}A_i : \forall i\in I \land x \in A_i\}$

Several things here don't make sense. $\forall i\in I$ is only half a statement ("for all i in I such that"), so it doesn't make sense to say that "and" something else. I also don't understand what you're trying to do.

Ok , so if $I = \{1,2,3,4,5,6,7,8,9\}$ then $I = \{n \in\mathbb Z\,|\,1\leq n\leq 9\}$?
...
Ok so we only use the number or symbol on top with we have $i=1$ at the bottom and not $i \in I$ ?

Yes to both questions.

What do you mean? I can't write it between {}?

You can, if you meant the singleton set whose only element is $\bigcup_{i=1}^3A_i$.

Ok , so B' doesn't really exist except in naive set theory?

I think that even in naive set theory, complements are always defined with respect to some set. You don't use the notation B' if it's not clear from the context what set $B\cup B'$ is.

 

[reenmachine]

First of all, you shouldn't have included the word "this", because it referred to the previous statement in my post. I assume that you meant that

x is an element of every $A_i$ with $i\in I$ $.\qquad(1)$​

means

$1,2,3 \in I$ and $x \in A_i$ $.\qquad(2)$​

Clearly (2) is very different from (1). (2) tells us that {1,2,3} is a subset of I. (1) tells us nothing about I. (2) tells us that $x\in A_i$. (What is i here?) (1) says that x is in all the A_i.

Several things here don't make sense. $\forall i\in I$ is only half a statement ("for all i in I such that"), so it doesn't make sense to say that "and" something else. I also don't understand what you're trying to do.

I was trying to write a set notation for $A_{1} \cup A_{2} \cup A_{3}$ with the knowledge that $I=\{1,2,3\}$

$$\{x \in \bigcup_{i=1}^3 A_i : \forall i~~\left(i\in I\ \Rightarrow x\in A_i\right)\}$$

Would that be a better set notation for $A_{1} \cup A_{2} \cup A_{3}$?

You can, if you meant the singleton set whose only element is $\bigcup_{i=1}^3A_i$.

what was I thinking...

I think that even in naive set theory, complements are always defined with respect to some set. You don't use the notation B' if it's not clear from the context what set $B\cup B'$ is.

ok thanks a lot man!

 

[Fredrik]

I was trying to write a set notation for $A_{1} \cup A_{2} \cup A_{3}$ with the knowledge that $I=\{1,2,3\}$

If you just want a notation for it, you can use $\{x\,|\,\exists i\in I~~ x\in A_i\}$.

Note that this wouldn't be a great way to define the notation $\bigcup_{i\in I}A_i$, since it it's not written in the way that's guaranteed to be "safe". Recall that $\{x\in y\,|\,P(x)\}$ is always OK, but $\{x\,|\,P(x)\}$ is sometimes not.

$$\{x \in \bigcup_{i=1}^3 A_i : \forall i~~\left(i\in I\ \Rightarrow x\in A_i\right)\}$$

Would that be a better set notation for $A_{1} \cup A_{2} \cup A_{3}$?

Since $A_{1} \cup A_{2} \cup A_{3}$ means the same thing as $\bigcup_{i=1}^3 A_i$, this wouldn't work as a definition, because of the problem of "using the definition while stating it".

If you meant this as a notation rather than as a definition, and you have previously specified that $I=\{1,2,3\}$, then you would have to change $\forall$ to $\exists$. If you do, you have a valid notation for $\bigcup_{i\in I}A_i$. But the $\in \bigcup_{i=1}^3 A_i$ before the colon looks pretty strange. If you want to put a set there in order to ensure that the notation is in the form that's guaranteed to be safe, it should be a set that's guaranteed to exist by the ZFC axioms and contains all the elements of all the $A_i$. (Edit: And it can't be the set we want to define).

 

[reenmachine]

If you just want a notation for it, you can use $\{x\,|\,\exists i\in I~~ x\in A_i\}$.

Note that this wouldn't be a great way to define the notation $\bigcup_{i\in I}A_i$, since it it's not written in the way that's guaranteed to be "safe". Recall that $\{x\in y\,|\,P(x)\}$ is always OK, but $\{x\,|\,P(x)\}$ is sometimes not.

Since $A_{1} \cup A_{2} \cup A_{3}$ means the same thing as $\bigcup_{i=1}^3 A_i$, this wouldn't work as a definition, because of the problem of "using the definition while stating it".

If you meant this as a notation rather than as a definition, and you have previously specified that $I=\{1,2,3\}$, then you would have to change $\forall$ to $\exists$. If you do, you have a valid notation for $\bigcup_{i\in I}A_i$. But the $\in \bigcup_{i=1}^3 A_i$ before the colon looks pretty strange. If you want to put a set there in order to ensure that the notation is in the form that's guaranteed to be safe, it should be a set that's guaranteed to exist by the ZFC axioms and contains all the elements of all the $A_i$. (Edit: And it can't be the set we want to define).

I did mean it as a notation , though in some ways I thought the definition should be a part of the notation , which I guess is wrong now.It's like my brain knows it's simple but yet I have some problems keeping the ''big picture'' of all these facts in my head.

I'm still having some confusion on when to use $\forall$ versus $\exists$ , which I guess remains a big problem.I'm trying to pinpoint where I'm confusing the two in my thought process.If $\forall i \in I \ x \in A_i$ it would mean that if $i=2$ , then $x \in A_i$ which is not true since if $i=2$ then $x \not \in A_1$ and $x \not \in A_3$.Is my observation correct? Then the $\exists$ instead of $\forall$ in $\exists i \in I \ x \in A_i$ makes the connection between $i \in I$ and $x \in A_i$ possible?

If $\{x\,|\,P(x)\}$ is not ok (sometimes) because of the Russell Paradox or similar problems , then we have $\{x\in y\,|\,P(x)\}$ which is always ok.That is very clear , yet if I try to use it with $A_1 \cup A_2 \cup A_3$ I'm confused about what $y$ could represent if $y$ doesn't equal the set we're trying to define.

$I=\{1,2,3\}$

$$\{x \in \ y : \exists i \in I \ x \in A_i\}$$
$$\{x \in \ ? : \exists i \in I \ x \in A_i\}$$

Suppose I try:
$$\{x \in A_i : \exists i \in I \ x \in A_i\}$$
Here $x \in A_i$ repeats itself , would this be an acceptable definition? (since $A_i$ isn't the set we're trying to define?)

Suppose:
$$\{x \in \bigcup_{i \in I} A_i : \exists i \in I \ x \in A_i\}$$

There the definition cannot be safe because I am using the set to define itself.I have a feeling I'm seeing all of this as more complicated than it really is , so in some ways that's comforting

Suppose:
$$\{x \in \bigcup A_i : \exists i \in I \ x \in A_i\}$$

does that even work? In that case the set we're trying to define would be a subset of $\bigcup A_i$.

To put it more precisely , to define set $A_1 \cup A_2 \cup A_3$ , $\{x \in y : \exists i \in I \ x \in A_i\}$ you need that $A_1 \cup A_2 \cup A_3 \subseteq y$ ?

thanks!

 

[Fredrik]

I'm still having some confusion on when to use $\forall$ versus $\exists$ , which I guess remains a big problem.I'm trying to pinpoint where I'm confusing the two in my thought process.If $\forall i \in I \ x \in A_i$ it would mean that if $i=2$ , then $x \in A_i$ which is not true since if $i=2$ then $x \not \in A_1$ and $x \not \in A_3$.Is my observation correct?

If you're considering an example such as the one with hair colors, then yes. Otherwise, it's possible that some x is an element of several of the $A_i$.

Then the $\exists$ instead of $\forall$ in $\exists i \in I \ x \in A_i$ makes the connection between $i \in I$ and $x \in A_i$ possible?

Not sure I understand this question. What I'm saying is that
\begin{align}
\{x\,|\,\exists i\in I~~ x\in A_i\} &=\bigcup_{i\in I}A_i\\
\{x\,|\,\forall i\in I~~ x\in A_i\} &=\bigcap_{i\in I}A_i
\end{align}

I'm confused about what $y$ could represent if $y$ doesn't equal the set we're trying to define.

That's understandable because I don't think there's a good answer. Maybe there is no better option than $y=\bigcup_{i\in I}A_i$. Books on set theory don't define unions with a notation like $\bigcup_{i\in I}A_i=\{x\in y\,|\,\exists i\in I~~ x\in A_i\}$ and a clever choice of y. They do it by referring to the "axiom of union", which says (this is a direct quote from Hrbacek & Jech): For any set S, there exists a set U such that $x\in U$ if and only if $x\in A$ for some $A\in S$.

If we apply this axiom to the set $S=\{A_1,A_2,A_3\}$, the $U$ that the axiom says exists is the set we denote by $\bigcup_{i\in I}A_i$. Since the axiom ensures that unions exist, there's no need to worry about the notation $\bigcup_{i\in I}A_i=\{x\,|\,\exists i\in I~~ x\in A_i\}$ not being in the "guaranteed safe" form. It also means that we can take y to be the set $\bigcup_{i\in I}A_i$ and write
$$\left\{x\in\bigcup_{i\in I}A_i~\Big|~\exists i\in I~~ x\in A_i\right\}$$ instead of $\bigcup_{i\in I}A_i$. It's not wrong, it just looks really weird. It's like writing $\{x\in A\,|\,x\in A\}$ instead of $A$. Since the statement $\exists i\in I~~ x\in A_i$ is true for all $x\in\bigcup_{i\in I}A_i$, we might as well write
$$\left\{x\in\bigcup_{i\in I}A_i~\Big|~x=x\right\}.$$ By the way, another notation for the set U is $\bigcup S$, which in our case means $\bigcup\{A_1,A_2,A_3\}$. This notation is sometimes useful (when we for some reason don't feel like defining an index set I and choosing a notation like $A_i$ for members of S), but the notations $\bigcup_{i\in I}A_i$ and $\bigcup_{i=1}^3 A_i$ are used much more often.

 

[reenmachine]

Not sure I understand this question. What I'm saying is that
\begin{align}
\{x\,|\,\exists i\in I~~ x\in A_i\} &=\bigcup_{i\in I}A_i\\
\{x\,|\,\forall i\in I~~ x\in A_i\} &=\bigcap_{i\in I}A_i
\end{align}

Oh ok , so by taking the second one and saying it in reverse it's $x \in A_i$ for all $i \in I$ meaning that $x$ will be in every $A_i$.

Just to be sure , in $\{x\,|\,\exists i\in I~~ x\in A_i\} =\bigcup_{i\in I}A_i$ , $x$ could be in a single $A_i$ or multiple ones is that correct?

To conclude with this , is $\{x\,|\,\exists i\in I~~ x\in A_i\}$ a notation of the form $\{x\,|\,P(x)\}$ ? If it's not , is it because of $i\in I$ ?

That's understandable because I don't think there's a good answer. Maybe there is no better option than $y=\bigcup_{i\in I}A_i$. Books on set theory don't define unions with a notation like $\bigcup_{i\in I}A_i=\{x\in y\,|\,\exists i\in I~~ x\in A_i\}$ and a clever choice of y. They do it by referring to the "axiom of union", which says (this is a direct quote from Hrbacek & Jech): For any set S, there exists a set U such that $x\in U$ if and only if $x\in A$ for some $A\in S$.

But isn't the set U guaranteed to be the exact same set as S? If U is every elements of S unionized , the result is the same no?

If we apply this axiom to the set $S=\{A_1,A_2,A_3\}$, the $U$ that the axiom says exists is the set we denote by $\bigcup_{i\in I}A_i$. Since the axiom ensures that unions exist, there's no need to worry about the notation $\bigcup_{i\in I}A_i=\{x\,|\,\exists i\in I~~ x\in A_i\}$ not being in the "guaranteed safe" form.

Doesn't $\{A_1,A_2,A_3\} = \bigcup_{i\in I}A_i$ ? (If $I = \{1,2,3\}$)

If so , doesn't $S = U$ ?

thanks man!

 

[Fredrik]

Oh ok , so by taking the second one and saying it in reverse it's $x \in A_i$ for all $i \in I$ meaning that $x$ will be in every $A_i$.

I wouldn't talk about x as if it's not a dummy variable here. But the statement $\forall i\in I~~ x\in A_i$ does of course mean that $x\in A_i$ for all $i\in I$.

Just to be sure , in $\{x\,|\,\exists i\in I~~ x\in A_i\} =\bigcup_{i\in I}A_i$ , $x$ could be in a single $A_i$ or multiple ones is that correct?

It doesn't really make sense to ask about a dummy variable, but it's true that an element of $\bigcup_{i\in I}A_i$ may belong to several of the $A_i$.

To conclude with this , is $\{x\,|\,\exists i\in I~~ x\in A_i\}$ a notation of the form $\{x\,|\,P(x)\}$ ?

It is, because $\exists i\in I~~ x\in A_i$ is a statement about x, just like P(x).

But isn't the set U guaranteed to be the exact same set as S?

No.
$$\bigcup\{\{1,2\},\{2,3\}\}=\{1,2,3\}\neq \{\{1,2\},\{2,3\}\}.$$

 

[reenmachine]

$$\bigcup\{\{1,2\},\{2,3\}\}=\{1,2,3\}\neq \{\{1,2\},\{2,3\}\}.$$

I want to make sure I understand this very clearly.

In your exemple , $S = \{\{1,2\},\{2,3\}\}$ and $\cup \ S = \{1,2,3\}$.

$1 \in \{1,2,3\}$ if and only if $1 \in A$ for some $A \in S$.

$A = \{1,2\} \ or \ \{2,3\}$ so $1 \in A$ for some $\{1,2\} \in \{\{1,2\},\{2,3\}\}$ is that a good way of understanding it?

Despite this , I'm still not sure what would happen if we took our previous exemple.

If $S = \{A_1 , A_2 , A_3 \}$ then what is $\cup$? Can $\cup = S$ sometimes , like in this case? Or is it that $\cup$ will be the set of all elements of $A_1$ , $A_2$ and $A_3$ , while $S$ is simply the set with three elements in the form of $A_i$ ?

thanks a lot!

 

[Fredrik]

In your exemple , $S = \{\{1,2\},\{2,3\}\}$ and $\cup \ S = \{1,2,3\}$.

Should be \bigcup instead of \cup here.

$1 \in \{1,2,3\}$ if and only if $1 \in A$ for some $A \in S$.

This is like saying "x=x if and only if x=x". It's not wrong, but it doesn't really say anything.

$A = \{1,2\} \ or \ \{2,3\}$

This statement is a bit strange since A is a dummy variable, but I think I know what you mean.

so $1 \in A$ for some $\{1,2\} \in \{\{1,2\},\{2,3\}\}$

This doesn't make sense. It should be ...for some $A \in \{\{1,2\},\{2,3\}\}$.

If $S = \{A_1 , A_2 , A_3 \}$ then what is $\cup$? Can $\cup = S$ sometimes , like in this case? Or is it that $\cup$ will be the set of all elements of $A_1$ , $A_2$ and $A_3$ , while $S$ is simply the set with three elements in the form of $A_i$ ?

This isn't really an example until you have specified what the $A_i$ sets are. So suppose that
\begin{align}
A_1&=\{1,3,5\}\\
A_2&=\{1,4,7\}\\
A_3&=\{1,5\}
\end{align}
What are the elements of the set
$$\bigcup \{A_1,A_2,A_3\} =A_1\cup A_2\cup A_2 =\bigcup_{i=1}^3 A_i =\bigcup_{i\in I} A_i?$$

Also, don't confuse the letter U with the symbol $\bigcup U$ in the Hrbacek & Jech definition.

 

[reenmachine]

This is like saying "x=x if and only if x=x". It's not wrong, but it doesn't really say anything.


This statement is a bit strange since A is a dummy variable, but I think I know what you mean.


This doesn't make sense. It should be ...for some $A \in \{\{1,2\},\{2,3\}\}$.

I was trying to introduce your exemple in your previous statement that:

For any set S, there exists a set U such that $x\in U$ if and only if $x\in A$ for some $A\in S$

$A \in \{\{1,2\},\{2,3\}\}$ means that A is either $\{1,2\}$ or $\{2,3\}$ no?

I used 1 for $x\in U$ , if and only if $1 \in A$ for some $A\in S$.Since $\{1,2\}$ could be A since $\{1,2\} \in S$ , then $1 \in U$.

This isn't really an example until you have specified what the $A_i$ sets are. So suppose that
\begin{align}
A_1&=\{1,3,5\}\\
A_2&=\{1,4,7\}\\
A_3&=\{1,5\}
\end{align}
What are the elements of the set
$$\bigcup \{A_1,A_2,A_3\} =A_1\cup A_2\cup A_2 =\bigcup_{i=1}^3 A_i =\bigcup_{i\in I} A_i?$$

$\{1,3,4,5,7\}$ ?

Also, don't confuse the letter U with the symbol $\bigcup U$ in the Hrbacek & Jech definition.

Is $\bigcup$ = $U$ or not? Is $\bigcup U$ an expression? What I mean is I thought it was $\bigcup S$ and not $\bigcup U$.I thought the set $U$ was already ''unionizing'' another set $S$?

thank you so much!

 

[Fredrik]

$A \in \{\{1,2\},\{2,3\}\}$ means that A is either $\{1,2\}$ or $\{2,3\}$ no?

Yes.

$\{1,3,4,5,7\}$ ?

Right. This is the U in Hrbacek & Jech's definition applied to the set $S=\{A_1,A_2,A_3\}=\{\{1,3,5\},\{1,4,7\},\{1,5\}\}$, and as you can see, S and U are very different.

Is $\bigcup$ = $U$ or not? Is $\bigcup U$ an expression? What I mean is I thought it was $\bigcup S$ and not $\bigcup U$.I thought the set $U$ was already ''unionizing'' another set $S$?

I don't remember the exact definition of "expression" from mathematical logic (I think there is an exact definition), but $\bigcup$ is a symbol that's used in the notation for unions. It doesn't denote a set. It looks like a U because that makes it easier to remember that it's part of the notation for unions. In Hrbacek & Jech's definition, U is a dummy variable. The definition says that given a set S, there exists a set U with a useful property. What I said is that this set is often denoted by $\bigcup S$.

 

[reenmachine]

Yes.Right. This is the U in Hrbacek & Jech's definition applied to the set $S=\{A_1,A_2,A_3\}=\{\{1,3,5\},\{1,4,7\},\{1,5\}\}$, and as you can see, S and U are very different.I don't remember the exact definition of "expression" from mathematical logic (I think there is an exact definition), but $\bigcup$ is a symbol that's used in the notation for unions. It doesn't denote a set. It looks like a U because that makes it easier to remember that it's part of the notation for unions. In Hrbacek & Jech's definition, U is a dummy variable. The definition says that given a set S, there exists a set U with a useful property. What I said is that this set is often denoted by $\bigcup S$.

All very clear! I'm very happy with the understanding I have of this section of the book of proof.The next section involves logic , which should be a lot of fun since we already talked about the basics earlier.

Thanks a lot Fredrik! Your help has been tremendously important for my progress.

 

[Fredrik]

Cool. You have certainly made a lot of progress. If you haven't already, you should examine the examples from sections 1.1-1.8 in the Book of Proof, and do some exercises from each of those sections. I recommend that you do enough exercises from each section that you feel like you could do all the others in the section if you wanted to.

 

[reenmachine]

I have some questions about the set $P(P(R^2))$. (here P = powerset , couldn't find how to write it in LaTeX.

Suppose we take (1,2) and (2,3) as elements of $R^2$.

$P(R^2)) = \{∅ , \ ... \ , \{(1,2)\} , \{(2,3)\} , \{(1,2),(2,3)\} , \ ... \ \}$

Would

$P(P(R^2)) = \begin{align}&\{∅ , \ ... \ , \{\{(1,2)\}\} , \{\{(2,3)\}\} , \{\{(1,2),(2,3)\}\} ,\\ &\{\{(2,3)\},\{(1,2),(2,3)\}\} , \{\{(1,2)\},\{(1,2),(2,3)\}\} , \{\{(1,2)\} , \{(2,3)\}\} , \\ &\{\{(1,2)\} , \{(2,3)\} , \{(1,2),(2,3)\}\} , \ ... \ \}\end{align}$ ?

I'm not sure if you understand what I'm trying to do , but basically I want the ''powersets of powersets'' concept to become very clear in my mind.

BTW , in the book of proof they denote the set $R^2$ with $\{(x,y) : x,y \in R\}$ , which is a notation of the form $\{x : p(x)\}$.It still confuses me a little bit when it is correct or not to use it.

About denoting the set $P(P(R^2))$ , would this be a possible notation:

$\{x : x \subset P(R^2)\}$

Here again I have the problem of leaving x alone on the left side.I could do:

$\{x \in ? : x \subseteq P(R^2)\}$
$\{x \in P(P(R^2)) : x \subseteq P(R^2)\}$

thanks !

EDIT: hmm seems like my long LaTeX sentence is ruining the size of the page.How can I fix this?

 

[Fredrik]

To fix the LaTeX, use an align environment. Click the quote button next to my post to see how I'm doing it. You can edit your post for 11 hours and 40 minutes. I like to use $\mathcal P$ (\mathcal P) for powersets.
\begin{align}
a &= b = c\\
&=d = e\\
&=f =g
\end{align} If you want to try to find $\mathcal P(\mathcal P(S))$ explicitly for some set S, I recommend that you consider a very simple S like S={1,2}, or even S={1}.

 

[reenmachine]

I recommend that you consider a very simple S like S={1,2}

$S=\{1,2\}$ $\mathcal P(S)=\{∅,\{1\},\{2\},\{1,2\}\}$$\mathcal P(\mathcal P(S))= \begin{align}&\{ \ ∅,\{\{1\}\},\{\{2\}\},\{\{1,2\}\},\{\{1\},\{1,2\}\},\\ &\{\{2\}\},\{1,2\}\},\{\{1\},\{2\}\},\\ &\{\{1\},\{2\},\{1,2\}\} \ \}\end{align}$

 

[Fredrik]

You have the right idea, but you forgot a few sets that involve the empty set. For example, when you listed all the subsets of $\mathcal P(S)$ with exactly 1 member, you left out $\{\emptyset\}$. When you listed the ones with exactly 2 members, you left out sets like $\{\emptyset,\{1\}\}$.

You should type the \begin{align} to the left of the code for your left-hand side, like I did in my example.

 

[reenmachine]

You have the right idea, but you forgot a few sets that involve the empty set. For example, when you listed all the subsets of $\mathcal P(S)$ with exactly 1 member, you left out $\{\emptyset\}$. When you listed the ones with exactly 2 members, you left out sets like $\{\emptyset,\{1\}\}$.

You should type the \begin{align} to the left of the code for your left-hand side, like I did in my example.

I was going to ask you about the ∅.Should ∅ be between brackets like {∅} when he's an element of a powerset? I thought ∅ was a special case that was always left alone as an element of every powerset , looks like I was wrong.

If $\{∅\}\in \mathcal P(S)$ , then $\{\{∅\}\} \in \mathcal P(P(S))$ but since $\{∅\}$ is a subset of every set , are $\{∅\}$ and $\{\{∅\}\}$ two separate elements of $\mathcal P(P(S))$?

For example, when you listed all the subsets of $\mathcal P(S)$ with exactly 1 member, you left out $\{\emptyset\}$

Here's what I wrote : $\mathcal P(S)=\{∅,\{1\},\{2\},\{1,2\}\}$

The ∅ is there , but not between brackets of his own.This is what you meant? If the brackets were missing , does the ∅ without brackets still stays there? (I think not)

thanks!

 

[Fredrik]

Your answer for $\mathcal P(S)$ was correct. We know that $\varnothing\subseteq S$, and that this implies that $\varnothing\in\mathcal P(S)$.

I was talking about your answer for $\mathcal P(\mathcal P(S))$, where your strategy appeared to be to first list all the subsets of $\mathcal P(S)$ with exactly 0 elements, then all subsets of $\mathcal P(S)$ with exactly 1 element,... This is a good strategy, but you left out a few sets.

Btw, there are two LaTeX codes for the empty set, \emptyset and \varnothing. It doesn't matter which one you use. The former looks like a zero with a slash through it, and the latter looks like a circle with a slash through it. I've been using \emptyset mostly, but I think I'm going to switch to \varnothing.

 

[reenmachine]

Your answer for $\mathcal P(S)$ was correct. We know that $\varnothing\subseteq S$, and that this implies that $\varnothing\in\mathcal P(S)$.

I was talking about your answer for $\mathcal P(\mathcal P(S))$, where your strategy appeared to be to first list all the subsets of $\mathcal P(S)$ with exactly 0 elements, then all subsets of $\mathcal P(S)$ with exactly 1 element,... This is a good strategy, but you left out a few sets.

Btw, there are two LaTeX codes for the empty set, \emptyset and \varnothing. It doesn't matter which one you use. The former looks like a zero with a slash through it, and the latter looks like a circle with a slash through it. I've been using \emptyset mostly, but I think I'm going to switch to \varnothing.

(EDIT: are you sure my answer is correct for $\mathcal P(S)$?)

Ok , I understand what you mean when you're saying that I forgot all the subsets of $\mathcal P(S)$ including $\varnothing$ in them.Basically I can reproduce all elements of $\mathcal P(\mathcal P(S))$ and add $\varnothing$ in it.Here's something I'm more confused about:

$\mathcal P(S)=\{\varnothing,\{1\},\{2\},\{1,2\}\}$ (for clearness)Since $\{\varnothing\} \subset \mathcal P(S)$ , then $\{\varnothing\} \in \mathcal P(\mathcal P(S))$.If you create the powerset $\mathcal P(\mathcal P(\mathcal P(S)))$ , are $\{\{\varnothing\}\} \subset \mathcal P(\mathcal P(S))$ and $\{\varnothing\} \subset \mathcal P(\mathcal P(S))$ ? Will both of them become separate elements of $\mathcal P(\mathcal P(\mathcal P(S)))$ ?

Last question , if $\{\{\varnothing\}\} \subset \mathcal P(\mathcal P(S))$ and $\{\varnothing\} \subset \mathcal P(\mathcal P(S))$ , then $\{\{\{\varnothing\}\},\{\varnothing\}\} \in \mathcal P(\mathcal P(\mathcal P(S)))$ ?

What I really want to know is if $\{\varnothing\}$ is an element of a powerset A because he was a subset of A , do we have to add a pair of brackets to ''transform'' him into a subset of powerset A?

If we do , $\{\{\varnothing\}\}$ will be a subset of power set A , but since $\{\varnothing\}$ is a subset of every set , then both $\{\{\varnothing\}\}$ and $\{\varnothing\}$ will be different subsets of power set A?

I might be severely overthinking all of this.Sorry if that's the case.

thanks a lot

 

[Fredrik]

(EDIT: are you sure my answer is correct for $\mathcal P(S)$?)

Yes. Since S={1,2} has two elements, a subset of S will have 0,1 or 2 elements.

S has one subset with exactly 0 elements: ∅
S has two subsets with exactly 1 element each: {1} and {2}
S has one subset with exactly 2 elements: {1,2}

Therefore P(S)={∅,{1},{2},{1,2}}.

Since $\{\varnothing\} \subset \mathcal P(S)$ , then $\{\varnothing\} \in \mathcal P(\mathcal P(S))$.If you create the powerset $\mathcal P(\mathcal P(\mathcal P(S)))$ , are $\{\{\varnothing\}\} \subset \mathcal P(\mathcal P(S))$ and $\{\varnothing\} \subset \mathcal P(\mathcal P(S))$ ? Will both of them become separate elements of $\mathcal P(\mathcal P(\mathcal P(S)))$ ?

Yes to all of that.

Last question , if $\{\{\varnothing\}\} \subset \mathcal P(\mathcal P(S))$ and $\{\varnothing\} \subset \mathcal P(\mathcal P(S))$ , then $\{\{\{\varnothing\}\},\{\varnothing\}\} \in \mathcal P(\mathcal P(\mathcal P(S)))$ ?

No to the conclusion. $\{\{\varnothing\}\}$ and $\{\varnothing\}$ will be elements of $\mathcal P(\mathcal P(\mathcal P(S)))$, so you can write $\{\{\varnothing\}\},\{\varnothing\}\in\mathcal P(\mathcal P(\mathcal P(S)))$ or $\{\{\{\varnothing\}\},\{\varnothing\}\}\subset \mathcal P(\mathcal P(\mathcal P(S)))$.

What I really want to know is if $\{\varnothing\}$ is an element of a powerset A because he was a subset of A , do we have to add a pair of brackets to ''transform'' him into a subset of powerset A?

For all sets x,y, $x\in y$ is equivalent to $\{x\}\subset y$.

For all sets A, the following statements are equivalent (either all true or all false).

1. ∅ is an element of A.
2. {∅} is a subset of A.
3. {∅} is an element of P(A).
4. {{∅}} is a subset of P(A).
5. {{∅}} is an element of P(P(A)).
...

If we do , $\{\{\varnothing\}\}$ will be a subset of power set A , but since $\{\varnothing\}$ is a subset of every set , then both $\{\{\varnothing\}\}$ and $\{\varnothing\}$ will be different subsets of power set A?

No, ∅ is a subset of every set, but {∅} is not. A set that has {∅} as a subset has ∅ as an element, and ∅ is certainly not an element of every set.

{{∅}} and {∅} are not the same. If they were, then since two sets are equal if and only if they have the same elements, we would have {∅}=∅. But the left-hand side is a set with 1 element, and the right-hand side is a set with 0 elements.

 

[reenmachine]

Yes. Since S={1,2} has two elements, a subset of S will have 0,1 or 2 elements.

S has one subset with exactly 0 elements: ∅
S has two subsets with exactly 1 element each: {1} and {2}
S has one subset with exactly 2 elements: {1,2}

Therefore P(S)={∅,{1},{2},{1,2}}.

Ok good

No to the conclusion. $\{\{\varnothing\}\}$ and $\{\varnothing\}$ will be elements of $\mathcal P(\mathcal P(\mathcal P(S)))$, so you can write $\{\{\varnothing\}\},\{\varnothing\}\in\mathcal P(\mathcal P(\mathcal P(S)))$ or $\{\{\{\varnothing\}\},\{\varnothing\}\}\subset \mathcal P(\mathcal P(\mathcal P(S)))$.

Ok , I'm pretty certain I either meant $\{\{\{\varnothing\}\},\{\varnothing\}\} \in \mathcal P(\mathcal P(\mathcal P(\mathcal P(S))))$ or $\{\{\{\varnothing\}\},\{\varnothing\}\} \subset \mathcal P(\mathcal P(\mathcal P(S)))$ so basically I was right in my head

For all sets x,y, $x\in y$ is equivalent to $\{x\}\subset y$.

Even though I knew it it's a nice statement to look at once in a while.

For all sets A, the following statements are equivalent (either all true or all false).

1. ∅ is an element of A.
2. {∅} is a subset of A.
3. {∅} is an element of P(A).
4. {{∅}} is a subset of P(A).
5. {{∅}} is an element of P(P(A)).
...

Ok , so in our case , since A isn't a powerset , then all the statements will be false since ∅ can only be an element of powersets.

The correct statements would be:
1. ∅ is a subset of A.
2. ∅ is an element of P(A).
3. {∅} is a subset of P(A).
4. {∅} is an element of P(P(A)).
5. {{∅}} is a subset of P(P(A)).
6. {{∅}} is an element of P(P(P(A))).

and to add to these statements:

7. ∅ is a subset of all sets.
8. ∅ is also an element of P(P(A)) and P(P(P(A))).
9. {∅} is also a subset of P(P(A)) and P(P(P(A))) because of that.

So basically , there's an accumulation of ∅ involved (even if they are between many brackets) the deeper you go into powersets? (and I know that they aren't the same thing but if we were to write many many powersets of powersets etc... the symbol ∅ would appear more and more the deeper you go , between brackets or not is that correct?)

No, ∅ is a subset of every set, but {∅} is not. A set that has {∅} as a subset has ∅ as an element, and ∅ is certainly not an element of every set.

{{∅}} and {∅} are not the same. If they were, then since two sets are equal if and only if they have the same elements, we would have {∅}=∅. But the left-hand side is a set with 1 element, and the right-hand side is a set with 0 elements

Ok that's good that you refreshed my mind with that.

BTW , \varnothing is definitely the best choice in LaTeX.

Thank you very much man!

 

[Fredrik]

Ok , so in our case , since A isn't a powerset , then all the statements will be false since ∅ can only be an element of powersets.

For all sets x with two elements or more, ∅ is an element of {∅,x}, and {∅,x} isn't the powerset of anything.

So basically , there's an accumulation of ∅ involved (even if they are between many brackets) the deeper you go into powersets? (and I know that they aren't the same thing but if we were to write many many powersets of powersets etc... the symbol ∅ would appear more and more the deeper you go , between brackets or not is that correct?)

Yes, that's right.

 

[reenmachine]

For all sets x with two elements or more, ∅ is an element of {∅,x}, and {∅,x} isn't the powerset of anything.
.

I don't understand this.I thought the empty set was generally not an element of sets.